document.write( "Question 408221: 1. The area of a rectangular room is 300 ft2. The perimeter is 70 ft. What are the dimensions of the room?
\n" ); document.write( "i tried to do this, but it doesnt work!!!!
\n" ); document.write( "100x3, 2x150, 1x300, 5x60, 6x50
\n" ); document.write( "thx! \n" ); document.write( "
Algebra.Com's Answer #287661 by ankor@dixie-net.com(22740)\"\" \"About 
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1. The area of a rectangular room is 300 ft2.
\n" ); document.write( "The perimeter is 70 ft. What are the dimensions of the room?
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\n" ); document.write( "The perimeter:
\n" ); document.write( "2L + 2W = 70
\n" ); document.write( "Simplify, divide by 2
\n" ); document.write( "L + W = 35
\n" ); document.write( "W = (35-L); use this form for substitution
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\n" ); document.write( "The area:
\n" ); document.write( "L * W = 300
\n" ); document.write( "replace W with (35-L)
\n" ); document.write( "L*(35-L) = 300
\n" ); document.write( "-L^2 + 35L = 300
\n" ); document.write( "-L^2 + 35L - 300 = 0
\n" ); document.write( "Easier to factor if L^2 is positive, multiply by -1
\n" ); document.write( "L^2 - 35L + 300 = 0
\n" ); document.write( "Factor this to
\n" ); document.write( "(L-15)(L-20) = 0
\n" ); document.write( "Two solutions
\n" ); document.write( "L = 15
\n" ); document.write( "L = 20 ft
\n" ); document.write( "choose the larger value for L, 20 ft, find W
\n" ); document.write( "W = 35 - 20
\n" ); document.write( "W = 15 ft is the width
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\n" ); document.write( "See if that is true
\n" ); document.write( "2(20) + 2(15) = 70 \n" ); document.write( "
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